is λ=h/mv the same equation as λ=hc/E ? Watch

>>MMM<<
Badges: 2
Rep:
?
#1
Report Thread starter 6 years ago
#1
if not when is each used?
0
reply
ibs2012
Badges: 6
Rep:
?
#2
Report 6 years ago
#2
Someone correct me if I'm wrong, but from what I remember:

λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc

At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E

So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.

I hope that helps.
2
reply
agostino981
Badges: 0
Rep:
?
#3
Report 6 years ago
#3
(Original post by ibs2012)
Someone correct me if I'm wrong, but from what I remember:

λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc

At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E

So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.

I hope that helps.
Photon is massless, so you can't really apply p=mv.

But, what you can do is apply E=h\nu and c=\lambda \nu, where \nu is the frequency and \lambda is the wavelength.

EDIT:
(Original post by >>MMM<<)
i am now confused so they are the same or not if not why and when do we use each?!
Photon is massless, so m=0 which gives you nothing if you use p=mv

What I did is to apply the equation for energy level E=h\nu and the equation relating the speed of wave, wavelength and frequency  v=f\lambda, because it is a photon, you have v=c, hence c=\nu \lambda

If you have something with mass, you can, of course, apply p=mv or p=\gamma mv for relativistic condition. However, you are considering photon, for momentum, you have p=\frac{h}{\lambda}.
3
reply
>>MMM<<
Badges: 2
Rep:
?
#4
Report Thread starter 6 years ago
#4
(Original post by agostino981)
Photon is massless, so you can't really apply p=mv.

But, what you can do is apply E=h\nu and c=\lambda \nu, where \nu is the frequency and \lambda is the wavelength.
i am now confused so they are the same or not if not why and when do we use each?!
0
reply
ibs2012
Badges: 6
Rep:
?
#5
Report 6 years ago
#5
(Original post by agostino981)
Photon is massless, so you can't really apply p=mv.

But, what you can do is apply E=h\nu and c=\lambda \nu, where \nu is the frequency and \lambda is the wavelength.
Ah yes, of course, thanks for the correction!
0
reply
ibs2012
Badges: 6
Rep:
?
#6
Report 6 years ago
#6
(Original post by >>MMM<<)
i am now confused so they are the same or not if not why and when do we use each?!
I'm assuming that your variables are m=mass,v=velocity?
In that case the only use I can find for λ=h/mv is when you are trying to find the de Broglie wavelength of larger objects, ie a car or person moving.
0
reply
>>MMM<<
Badges: 2
Rep:
?
#7
Report Thread starter 6 years ago
#7
(Original post by ibs2012)
I'm assuming that your variables are m=mass,v=velocity?
In that case the only use I can find for λ=h/mv is when you are trying to find the de Broglie wavelength of larger objects, ie a car or person moving.
how is that possible? i have read in my book that λ=h/mv can be used in any situation?
0
reply
agostino981
Badges: 0
Rep:
?
#8
Report 6 years ago
#8
(Original post by >>MMM<<)
how is that possible? i have read in my book that λ=h/mv can be used in any situation?
Except massless or static object.
0
reply
bbcs2k9
Badges: 3
Rep:
?
#9
Report 6 years ago
#9
(Original post by ibs2012)
Someone correct me if I'm wrong, but from what I remember:

λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc

At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E

So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.

I hope that helps.
λ=h/mv is known as the Broglie wavelength and you use it when you are asked to find the associated wavelength of a particle such as an electron with a certain speed.

λ=hc/E is the wavelength of a photon with a certain energy. You usually see the equation in this form: E=hc/λ. and they will usually ask to find the wavelength energy given a certain wavelength.

I dont think they are they same (they might be, but I dont' know). But generally in my course (OCR), λ=h/mv is for particles and E=hc/λ is for photons of light.
0
reply
>>MMM<<
Badges: 2
Rep:
?
#10
Report Thread starter 6 years ago
#10
(Original post by bbcs2k9)
λ=h/mv is known as the Broglie wavelength and you use it when you are asked to find the associated wavelength of a particle such as an electron with a certain speed.

λ=hc/E is the wavelength of a photon with a certain energy. You usually see the equation in this form: E=hc/λ. and they will usually ask to find the wavelength energy given a certain wavelength.

I dont think they are they same (they might be, but I dont' know). But generally in my course (OCR), λ=h/mv is for particles and E=hc/λ is for photons of light.
that was the answer i was looking forward to but the problem is that my book says that a beam of moving electrons has associated de broglie wavelength (where at high speed you have to take into account relativistic effects) is approx λ≃hc/E can you please tell me what does that mean?
0
reply
theterminator
Badges: 1
Rep:
?
#11
Report 6 years ago
#11
(Original post by >>MMM<<)
that was the answer i was looking forward to but the problem is that my book says that a beam of moving electrons has associated de broglie wavelength (where at high speed you have to take into account relativistic effects) is approx λ≃hc/E can you please tell me what does that mean?
If you want to calculate the relativistic de broglie wavelength of a particle of rest mass m_{0} then the equation becomes:

\displaystyle {\lambda}_{de-Broglie} = \frac{h}{vcm_{0}} \displaystyle \sqrt_{c^2 - v^2}
0
reply
>>MMM<<
Badges: 2
Rep:
?
#12
Report Thread starter 6 years ago
#12
(Original post by theterminator)
If you want to calculate the relativistic de broglie wavelength of a particle of rest mass m_{0} then the equation becomes:

\displaystyle {\lambda}_{de-Broglie} = \frac{h}{vcm_{0}} \displaystyle \sqrt_{c^2 - v^2}
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!
0
reply
theterminator
Badges: 1
Rep:
?
#13
Report 6 years ago
#13
(Original post by >>MMM<<)
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!
I think you may be getting confused with the derivation of the de Broglie equation from above. The above derivation is the one taught but isn't actually the correct way of deriving the de Broglie wavelength. The correct derivation requires more fundamental special relativity but it makes the clear distinction between photons and electrons. Something you should know about the de Broglie wavelength is that it is a relativistic effect, and E=mc^2 is an equation for something that has zero velocity in the frame of reference being used.
0
reply
Pangol
Badges: 12
Rep:
?
#14
Report 6 years ago
#14
(Original post by >>MMM<<)
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!
Look again at bbcs2k9's post above. E=hf (or E=hc/λ, they are the same equation) is for photons. If you wnat to find the deBroglie wavelength of a particle, you need to use λ=h/mv.
0
reply
agostino981
Badges: 0
Rep:
?
#15
Report 6 years ago
#15
(Original post by >>MMM<<)
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!
Nope, \lambda = \frac{hc}{E} is for photons only.

If you want to find the de-Broglie wavelength of electrons, you need \lambda=\frac{h}{p}.

If you seek a non-relativistic result, you apply p=mv, if you seek a relativistic result, you apply p=\gamma m v , where \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

A pretty obvious reason why you can't use \lambda=\frac{hc}{E} is because electrons cannot reach light speed.
1
reply
>>MMM<<
Badges: 2
Rep:
?
#16
Report Thread starter 6 years ago
#16
(Original post by agostino981)
Nope, \lambda = \frac{hc}{E} is for photons only.

If you want to find the de-Broglie wavelength of electrons, you need \lambda=\frac{h}{p}.

If you seek a non-relativistic result, you apply p=mv, if you seek a relativistic result, you apply p=\gamma m v , where \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

A pretty obvious reason why you can't use \lambda=\frac{hc}{E} is because electrons cannot reach light speed.
so your basically saying λ=h/mv is only for particles and λ=hc/E only for photons and the book has some kind of mistake in it?
0
reply
>>MMM<<
Badges: 2
Rep:
?
#17
Report Thread starter 6 years ago
#17
(Original post by Pangol)
Look again at bbcs2k9's post above. E=hf (or E=hc/λ, they are the same equation) is for photons. If you wnat to find the deBroglie wavelength of a particle, you need to use λ=h/mv.
soo your saying λ=h/mv is can only for particles and λ=hc/E can only for photons and the book has some kind of mistake in it?
0
reply
>>MMM<<
Badges: 2
Rep:
?
#18
Report Thread starter 6 years ago
#18
(Original post by theterminator)
I think you may be getting confused with the derivation of the de Broglie equation from above. The above derivation is the one taught but isn't actually the correct way of deriving the de Broglie wavelength. The correct derivation requires more fundamental special relativity but it makes the clear distinction between photons and electrons. Something you should know about the de Broglie wavelength is that it is a relativistic effect, and E=mc^2 is an equation for something that has zero velocity in the frame of reference being used.
so λ=hc/E can't be used for particles??
0
reply
Pangol
Badges: 12
Rep:
?
#19
Report 6 years ago
#19
(Original post by >>MMM<<)
soo your saying λ=h/mv is can only for particles and λ=hc/E can only for photons and the book has some kind of mistake in it?
Yes (although I'd have to see exactly what the book said before saying categorically that it is wrong).
1
reply
theterminator
Badges: 1
Rep:
?
#20
Report 6 years ago
#20
(Original post by >>MMM<<)
so λ=hc/E can't be used for particles??
It cannot be used on particles with mass; i.e. It can only be used on photons/light (because photons are massless).
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts

University open days

  • Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 31 Jul '19
  • Staffordshire University
    Postgraduate open event - Stoke-on-Trent campus Postgraduate
    Wed, 7 Aug '19
  • University of Derby
    Foundation Open Event Further education
    Wed, 7 Aug '19

Are cats selfish

Yes (139)
61.23%
No (88)
38.77%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise